The diagram shows a graph. For a uniform metallic wire, what could the graph not represent?

Question 17

The diagram shows a graph.

For a uniform metallic wire, what could the graph not represent?

y-axis                          x-axis

A          current                         potential difference

B          resistance                    length

C         resistance                    temperature in °C

D         potential difference     current

Reference: Past Exam Paper – November 2015 Paper 11 Q31

Solution:

Answer: C.

A uniform metallic wire obeys Ohm’s law (V = IR). So, a current-p.d. graph and a p.d.-current graph would be straight line. [A and D incorrect]

Resistance of wire: R = ρL / A

So, a resistance-length graph would also be a straight line. [B incorrect]

A graph of resistance against temperature will not be a straight line as the resistance of the metal wire increases with temperature. The higher the temperature, the greater the increase in temperature. So, the graph would be a curve, not a straight line.

• 9702 November 2017 Paper 12 Worked Solution | A-Level Physics

1 D                  11 D                21 A                31 A

2 A                  12 B                22 D                32 B

3 C                  13 D                23 D                33 D

4 D                  14 A                24 B                34 C

5 C                  15 B                25 C                35 A

6 B                  16 A                26 D                36 B

7 B                  17 D                27 D                37 B

8 D                  18 A                28 C                38 A

9 C                  19 B                29 C                39 A

10 B                20 D                30 D                40 C

6 - Ans: B.

A hot-air balloon is moving vertically upwards with a constant speed of 3.00 m s-1.

12 - Ans: B.
The diagram shows a solid cube with weight W and sides of length L. It is supported at rest by a frictionless spindle that passes through the centres of two opposite vertical faces.


14 - Ans: A.
Two rigid steel beams XY and YZ are fixed at their lower ends and are hinged at Y. Each beam is inclined at 50 to the horizontal, as shown.



15 - Ans: B.
Air is trapped inside a glass bulb which is immersed in water and attached to a U-tube containing mercury.

21 - Ans: A.

A rectangular block of steel supporting a very large component of a bridge has a height of 15 cm and a cross-section of 20 cm × 12 cm.

32 - Ans: B.
An electron moves between two points X and Y in a uniform electric field, as shown.   The distance between X and Y is 4.0 cm and the line XY is at an angle of 60 to the direction of the field.



37 - Ans: B.
The circuit shown contains a resistor S that is neither in series nor in parallel with the other resistors.

The Young modulus of steel is twice that of copper. A 50 cm length of copper wire of diameter 2.0 mm is joined to a 50 cm length of steel wire of diameter 1.0 mm, making a combination wire of length 1.0 m, as shown.

Question 9

The Young modulus of steel is twice that of copper.

A 50 cm length of copper wire of diameter 2.0 mm is joined to a 50 cm length of steel wire of diameter 1.0 mm, making a combination wire of length 1.0 m, as shown.

The combination wire is stretched by a weight added to its end. Both the copper and the steel wires obey Hooke’s law.

What is the ratio

extension of steel wire      ?

extension of copper wire       

A 4                   B 2                   C 1                  D 0.5

Reference: Past Exam Paper – March 2016 Paper 12 Q19

Solution:

Answer: B.

Young modulus E = Stress / Strain

E = FL / Ae

The length L of both the copper wire and the steel wire is 50 cm each.

 

A weight is attached at the end of the combination wire. This causes a force F on each wire.

Cross-sectional area A = πd² / 4

The cross-sectional area is proportional to the square of the diameter (d²). Since Young modulus E = FL / Ae, we can say that the Young modulus is proportional to 1/d² (since the π/4 in the area is a constant).

The Young modulus of steel is twice that of copper.

E_steel = 2 × E_copper

F × 50 / (d_steel² ×e_steel) = 2 × F × 50 / (d_copper² ×e_copper)

1 / (d_steel² ×e_steel) = 2 / (d_copper² ×e_copper)

e_steel / e_copper = d_copper² / 2×d_steel² = 2.0² / (2×1.0²) 

e_steel / e_copper = 4 / 2 = 2